Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $n = \dfrac{k - 5}{-9k^2 + 9k} \div \dfrac{k + 6}{k^2 + 5k - 6} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{k - 5}{-9k^2 + 9k} \times \dfrac{k^2 + 5k - 6}{k + 6} $ First factor the quadratic. $n = \dfrac{k - 5}{-9k^2 + 9k} \times \dfrac{(k - 1)(k + 6)}{k + 6} $ Then factor out any other terms. $n = \dfrac{k - 5}{-9k(k - 1)} \times \dfrac{(k - 1)(k + 6)}{k + 6} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (k - 5) \times (k - 1)(k + 6) } { -9k(k - 1) \times (k + 6) } $ $n = \dfrac{ (k - 5)(k - 1)(k + 6)}{ -9k(k - 1)(k + 6)} $ Notice that $(k + 6)$ and $(k - 1)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ (k - 5)\cancel{(k - 1)}(k + 6)}{ -9k\cancel{(k - 1)}(k + 6)} $ We are dividing by $k - 1$ , so $k - 1 \neq 0$ Therefore, $k \neq 1$ $n = \dfrac{ (k - 5)\cancel{(k - 1)}\cancel{(k + 6)}}{ -9k\cancel{(k - 1)}\cancel{(k + 6)}} $ We are dividing by $k + 6$ , so $k + 6 \neq 0$ Therefore, $k \neq -6$ $n = \dfrac{k - 5}{-9k} $ $n = \dfrac{-(k - 5)}{9k} ; \space k \neq 1 ; \space k \neq -6 $